ĐK: $x \ge -2$
Ta có:
$$4\sqrt{x+2}+\sqrt{22-3x}=x^2+8 \\\iff 4(\sqrt{x+2}-1)+(\sqrt{22-3x}-5)+(1-x^2)=0 \\\iff \dfrac{4(x+1)}{\sqrt{x+2}+1}+\dfrac{-3x-3}{\sqrt{22-3x}+5}+(1+x)(1-x)=0 \\\iff (x+1)(\dfrac{4}{\sqrt{x+2}+1} -\dfrac{3}{\sqrt{22-3x}+5}+1-x)=0 \\\iff \left\{\begin{matrix} x=-1 \\ \dfrac{4}{\sqrt{x+2}+1} -\dfrac{3}{\sqrt{22-3x}+5}+1-x=0 \end{matrix}\right. $$